A force F of magnitude 2x^3 is applied to stop a particle moving with an initial velocity of v0. The particle travels from x=0 t o x= D in coming to a stop after the force F is applied. The work done by F is:a) 1/2 D^3b) -D^4c) D^4d) 2D^3e) -D^3f) -1/2 D^4g) 1/2D^4h) -1/2D^3
1 answer:
Answer:
Explanation:
Given that
F=2x³
Work is given as
The range of x is from x=0 to x=D
W=-∫f(x)dx
Then,
W=-∫2x³dx from x=0 to x=D
W=- 2x⁴/4 from x=0 to x=D
W=-2(D⁴/4-0/4)
W=-D⁴/2
W=1/2D⁴
The correct answer is F
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